8Lubin–Tate theory
III Local Fields
8.3 Lubin–Tate extensions
We now use the Lubin–Tate modules to do things. As before, we fixed an
algebraic closure
¯
K of K. We let
¯
m = m
¯
K
be the maximal ideal in O
¯
K
.
Proposition.
If
F
is a formal
O
K
module, then
¯
m
becomes a (genuine)
O
K
module under the operations +
F
and ·
x +
F
y = F(x, y)
a · x = [a]
F
(x)
for all x, y ∈
¯
m and a ∈ O
K
.
We denote this
¯
m
F
.
This isn’t exactly immediate, because
¯
K
need not be complete. However,
this is not a problem as each multiplication given by
F
only involves finitely
many things (namely two of them).
Proof.
If
x, y ∈
¯
m
, then
F
(
x, y
) is a series in
K
(
x, y
)
⊆
¯
K
. Since
K
(
x, y
) is
a finite extension, we know it is complete. Since the terms in the sum have
absolute value
<
1 and
→
0, we know it converges to an element in
m
K(x,y)
⊆
¯
m
.
The rest then essentially follows from definition.
To prove local class field theory, we want to find elements with an
U
K
/U
(n)
K
action for each
n
, or equivalently elements with an
O
K
/O
(n)
K
action. Note that
the first quotient is a quotient of groups, while the second quotient is a quotient
of a ring by an ideal. So it is natural to consider the following elements:
Definition
(
π
n
division points)
.
Let
F
be a Lubin–Tate
O
K
module for
π
. Let
n ≥ 1. The group F (n) of π
n
division points of F is defined to be
F (n) = {x ∈
¯
m
F
 [π
n
]
F
x = 0} = ker([π
n
]
F
).
This is a group under the operation given by F , and is indeed an O
K
module.
Example. Let F =
ˆ
G
m
, K = Q
p
and π = p. Then for x ∈
¯
m
ˆ
G
m
, we have
p
n
· x = (1 + x)
p
n
− 1.
So we know
ˆ
G
m
(n) = {ζ
i
p
n
− 1  i = 0, 1, ··· , p
n
− 1},
where ζ
p
n
∈
¯
Q
p
is the primitive p
n
th root of unity.
So
ˆ
G
m
(n) generates Q
p
(ζ
p
n
).
To prove this does what we want, we need the following lemma:
Lemma. Let e(X) = X
q
+ πX. We let
f
n
(X) = (e ◦ ··· ◦ e)
 {z }
n times
(X).
Then f
n
has no repeated roots. Here we take f
0
to be the identity function.
Proof.
Let
x ∈
¯
K
. We claim that if
f
i
(
x
)
 <
1 for
i
= 0
, ··· , n −
1, then
f
0
n
(X) 6= 0.
We proceed by induction on n.
(i) When n = 1, we assume x < 1. Then
f
0
1
(x) = e
0
(x) = qx
q−1
+ π = π
1 +
q
π
x
q−1
6= 0,
since we know
q
π
has absolute value
≤
1 (
q
vanishes in
k
K
, so
q/π
lives in
O
K
), and x
q−1
has absolute value < 1.
(ii) in the induction step, we have
f
0
n+1
(x) = (qf
n
(x)
q−1
+ π)f
0
n
(x) = π
1 +
q
π
f
n
(x)
q−1
f
0
n
(x).
By induction hypothesis, we know
f
0
n
(
x
)
6
= 0, and by assumption
f
n
(
x
)
 <
1. So the same argument works.
We now prove the lemma. We assume that
f
n
(
x
) = 0. We want to show that
f
i
(x) < 1 for all i = 0, ··· , n − 1. By induction, we have
f
n
(x) = x
q
n
+ πg
n
(x)
for some
g
n
(
x
)
∈ O
K
[
x
]. It follows that if
f
n
(
x
) = 0, then
x <
1. So
f
i
(
x
)
 <
1
for all i. So f
0
n
(x) 6= 0.
The point of the lemma is to prove the following proposition:
Proposition. F
(
n
) is a free
O
K
/π
n
O
K
module of rank 1. In particular, it has
q
n
elements.
Proof. By definition, we know
π
n
· F (n) = 0.
So F (n) is indeed an O
K
/π
n
O
K
module.
To prove that it is free of rank 1, we note that all Lubin–Tate modules
for
π
are isomorphic. This implies that all the honest
O
K
modules
F
(
n
) are
isomorphic. We choose
F
=
F
e
, where
e
=
X
q
+
πX
. Then
F
(
n
) consists
of the roots of the polynomial
f
n
=
e
n
(
X
), which is of degree
q
n
and has no
repeated roots. So
F
(
n
)

=
q
n
. To show that it is actually the right thing, if
λ
n
∈ F (n) \ F (n −1), then we have a homomorphism
O
K
→ F (n)
given by
A 7→ a · λ
n
. Its kernel is
π
n
O
K
by our choice of
λ
n
. By counting, we
get an O
K
module isomorphism
O
K
π
n
O
K
→ F (n)
as desired.
Corollary. We have isomorphisms
O
K
π
n
O
K
∼
=
End
O
K
(F (n))
U
K
U
(n)
K
∼
=
Aut
O
K
(F (n)).
Given a Lubin–Tate O
K
module F for π, we consider
L
n,π
= L
n
= K(F (n)),
which is the field of
π
n
division points of
F
. From the inclusions
F
(
n
)
⊆ F
(
n
+1)
for all n, we obtain a corresponding inclusion of fields
L
n
⊆ L
n+1
.
The field
L
n
depends only in
π
, and not on
F
. To see this, we let
G
be another
Lubin–Tate O
K
module, and let f : F → G be an isomorphism. Then
G(n) = f(F (n)) ⊆ K(F (n))
since the coefficients of f lie in K. So we know
K(G(n)) ⊆ K(F (n)).
By symmetry, we must have equality.
Theorem. L
n
/K
is a totally ramified abelian extension of degree
q
n−1
(
q −
1)
with Galois group
Gal(L
n
/K)
∼
=
Aut
O
K
(F (n))
∼
=
U
K
U
(n)
K
.
Explicitly, for any σ ∈ Gal(L
n
/K), there is a unique u ∈ U
K
/U
(n)
K
such that
σ(λ) = [u]
F
(λ)
for all λ ∈ F (n). Under this isomorphism, for m ≥ n, we have
Gal(L
m
/L
n
)
∼
=
U
(n)
K
U
(m)
K
.
Moreover, if F = F
e
, where
e(X) = X
q
+ π(a
q−1
π
q−1
+ ··· + a
2
X
2
) + πX,
and λ
n
∈ F (n) \ F (n −1), then λ
n
is a uniformizer of L
n
and
φ
n
(X) =
e
n
(X)
e
n−1
(X)
= X
q
n−1
(q−1)
+ ··· + π
is the minimal polynomial of λ
n
. In particular,
N
L
n
/K
(−λ
n
) = π.
Proof. Consider a Lubin–Tate polynomial
e(X) = x
q
+ π(a
q−1
X
q−1
+ ··· + a
2
X
2
) + πX.
We set F = F
e
. Then
φ
n
(X) =
e
n
(X)
e
n−1
(X)
= (e
n−1
(X))
q−1
+ π(a
q
1
e
n−1
(X)
q−2
+ ···+ a
2
e
n−1
(X)) + π
is an Eisenstein polynomial of degree
q
n−1
(
q −
1) by starting at it long enough.
So if
λ
n
∈ F
(
n
)
\ F
(
n −
1), then
λ
n
is a root of
φ
n
(
x
), so
K
(
λ
n
)
/K
is totally
ramified of degree q
n−1
(q −1), and λ
n
is a uniformizer, and
N
K(λ
n
)/K
(−λ
n
) = π
as the norm is just the constant coefficient of the minimal polynomial.
Now let
σ ∈ Gal
(
L
n
/K
). Then
σ
induces a permutation of
F
(
n
), as these
are the roots of e
n
(X), which is in fact O
K
linear, i.e.
σ(x) +
F
σ(y) = F (σ(x), σ(y)) = σ(F (x, y)) = σ(x +
F
y)
σ(a · x) = σ([a]
F
(x)) = [a]
F
(σ(x)) = a · σ(x)
for all x, y ∈ m
L
n
and a ∈ O
K
.
So we have an injection of groups
Gal(L
n
/K) → Aut
O
K
(F (n)) =
U
K
U
(n)
K
But we know
U
K
U
(n)
K
= q
n−1
(q −1) = [K(λ
n
) : K] ≤ [L
n
: K] = Gal(L
n
/K).
So we must have equality throughout, the above map is an isomorphism, and
K(λ
n
) = L
n
.
It is clear from the construction of the isomorphism that for
m ≥ n
, the
diagram
Gal(L
m
/K) U
K
/U
(m)
K
Gal(L
n
/K) U
K
/U
(n)
K
∼
restriction
quotient
∼
commutes. So the isomorphism
Gal(L
m
/L
n
)
∼
=
U
(m)
K
U
(n)
K
follows by looking at the kernels.
Example. In the case where K = Q
p
and π = p, recall that
ˆ
G
m
(n) = {ζ
i
p
n
− 1  i = 0, ··· , p
n−1
− 1},
where ζ
p
n
is the principal p
n
th root of unity. The theorem then gives
Gal(Q
p
(ζ
p
n
)/Q
p
)
∼
=
(Z/p
n
)
×
given by if a ∈ Z
≥0
and (a, p) = 1, then
σ
a
(ζ
i
p
n
− 1) = [a]
ˆ
G
m
(n)
(ζ
i
p
n
− 1) = (1 + (ζ
i
p
n
− 1))
a
− 1 = ζ
ai
p
n
− 1.
This agrees with the isomorphism we previously constructed.
Back to the general situation, setting
L
∞
=
∞
[
n=1
L
n
,
we know L
∞
/K is Galois, and we have isomorphisms
Gal(L
∞
/K) lim
←−
Gal(L
n
/K) lim
←−
n
U
K
/U
(n)
K
∼
=
U
K
σ (σ
L
n
)
n
∼ ∼
This map will be the inverse of the Artin map restricted to L
∞
.
To complete the proof of Artin reciprocity, we need to use the following
theorem without proof:
Theorem (Generalized local KroneckerWeber theorem). We have
K
ab
= K
ur
L
∞
(for any π).
Comments on the proof.
One can prove this from the HasseArf theorem, which
states that in an abelian extension, the jumps in the upper ramification groups
occur only at integer values. This, together with the calculation of ramification
groups done later, easily implies the theorem. Essentially,
L
∞
maxed out all
possible jumps of the upper ramification groups. However, the HasseArf theorem
is difficult to prove.
Another approach is to prove the existence of the Artin map using other
techniques (e.g. Galois cohomology). Consideration of the norm group (cf. the
next theorem) then implies the theorem. The content of this section then
becomes an explicit construction of a certain family of abelian extensions.
We can characterize the norm group by
Theorem. We have
N(L
n
/K) = hπi × U
(n)
k
.
Comments on the proof.
This can be done by defining Coleman operators, which
are power series representations of the norm. Alternatively, assuming the
description of the local Artin map given below and local Artin reciprocity,
U
(n)
k
is in the kernel of
Art
L
n
, so
hπi × U
(n)
k
⊆ N
(
L
n
/K
). The result follows by
comparing order.
We can then construct the Artin map as follows:
Theorem.
Let
K
be a local field. Then we have an isomorphism
Art
:
K
×
→
W (K
ab
/K) given by the composition
K
×
W (K
ab
/K)
hπi × U
K
Frob
Z
K
× Gal(L
∞
/K)
∼
Art
∼
where the bottom map is given by (π
m
, u) 7→ (Frob
m
K
, σ
u
−1
), where
σ
u
(λ) = [u]
F
(λ)
for all λ ∈
S
∞
n=1
F (n).
The inverse shows up in the proof to make sure the map defined above is
independent of the choice of uniformizer. We will not prove this, nor that the
map obtained has the desired properties. Instead, we will end the course by
computing the higher ramification groups of these extensions.
Theorem. We have
G
s
(L
n
/K) =
Gal(L
n
/K) −1 ≤ s ≤ 0
Gal(L
n
/L
k
) q
k−1
− 1 < s ≤ q
k
− 1, 1 ≤ k ≤ n − 1
1 s > q
n−1
Proof. The case for −1 ≤ s ≤ 0 is clear.
For 0 ≤ s ≤ 1 (which we may wlog is actually 1), we know that
Gal(L
n
/L
k
)
∼
=
U
(k)
K
/U
(n)
K
under the isomorphism
Gal
(
L
n
/K
)
∼
=
U
K
/U
(n)
K
. On the other hand, we know
G
1
(L
n
/K) is the Sylow psubgroup of Gal(L
n
/K). So we must have
G
1
(L
n
/K)
∼
=
U
(1)
K
/U
(n)
K
.
So we know that
G
1
(
L
n
/K
) =
Gal
(
L
n
/L
1
). Thus we know that
G
s
(
L
n
/K
) =
Gal(L
n
/K) for 0 < s ≤ 1.
We now let σ = σ
u
∈ G
1
(L
n
/K) and u ∈ U
(1)
K
/U
(n)
K
. We write
u = 1 + επ
k
for some
ε ∈ U
K
and some
k
=
k
(
u
)
≥
1. Since
σ
is not the identity, we know
k < n. We claim that
i
L
n
/K
(σ) = v
L
n
(σ(λ) − λ) = q
k
.
Indeed, we let
λ ∈ F
(
n
)
\ F
(
n −
1), where
F
is a choice of Lubin–Tate module
for π. Then λ is a uniformizer of L
n
and O
L
n
= O
K
[λ]. We can compute
σ
u
(λ) = [u]
F
(λ)
= [1 + επ
k
]
F
(λ)
= F (λ, [επ
k
]
F
(λ))
Now we can write
[επ
k
]
F
(λ) = [ε]
F
([π
k
]
F
(λ)) ∈ F (n − k) \ F (n − k − 1),
since [
ε
]
F
is invertible, and applying [
π
n−k
]
F
to [
π
k
]
F
(
λ
) kills it, but applying
[π
n−k−1
]
F
gives [π
n−1
]
F
, which does not kill.
So we know [
επ
k
]
F
(
λ
) is a uniformizer of
L
n−k
. Since
L
n
/L
n−k
is totally
ramified of degree q
k
, we can find ε
0
∈ O
×
L
n
such that
[επ
k
]
F
(λ) = ε
0
λ
q
k
Recall that F (X, 0) = X and F (0, Y ) = Y . So we can write
F (X, Y ) = X + Y + XY G(X, Y ),
where G(X, Y ) ∈ O
K
[[X, Y ]]. So we have
σ(λ) − λ = F (λ, [επ
k
]
F
(λ)) − λ
= F (λ, ε
0
λ
q
k
) − λ
= λ + ε
0
λ
q
k
+ ε
0
λ
q
k
+1
G(λ, ε
0
λ
q
k
) − λ
= ε
0
λ
q
k
+ ε
0
λ
q
k
+1
G(λ, ε
0
λ
q
k
).
In terms of valuation, the first term is the dominating term, and
i
L
n
/K
(σ) = v
L
n
(σ(λ) − λ) = q
k
So we know
i
L
n
/K
(σ
k
) ≥ s + 1 ⇔ q
k(u)
− 1 ≥ s.
So we know
G
s
(L
n
/K) = {σ
K
∈ G
1
(L
n
/K) : q
k(u)
− 1 ≥ s} = Gal(L
n
/L
k
),
where q
k−1
− 1 < s ≤ q
k
− 1 for k = 1, ··· , n − 1, and 1 if s > q
n−1
= 1.
Corollary. We have
G
t
(L
n
/K) =
Gal(L
n
/K) −1 ≤ t ≤ 0
Gal(L
n
/L
k
) k − 1 < t ≤ k, k = 1, ··· , n − 1
1 t > n − 1
In other words, we have
G
t
(L
n
/K) =
(
Gal(L
n
/L
dte
) −1 ≤ t ≤ n − 1
1 t > n − 1
,
where we set L
0
= K.
Once again, the numbering is a bit more civilized in the upper numbering.
Proof. We have to compute the integral of
1
(G
0
(L
n
/K) : G
x
(L
n
/K)
.
We again plot this out
1
q−1
1
q(q−1)
1
q
2
(q−1)
q − 1
q
2
− 1 q
3
− 1
So by the same computation as the ones we did last time, we find that
η
L
n
/K
(s) =
s −1 ≤ s ≤ 0
(k − 1) +
s−(q
k−1
−1)
q
k−1
(q−1)
q
k−1
− 1 ≤ s ≤ q
k
− 1, k = 1, ··· , n − 1
(n − 1) +
s−(q
n−1
−1)
q
n−1
(q−1)
s > q
n−1
− 1.
Inverting this, we find that
ψ
L
n
/K
=
t −1 ≤ t ≤ 0
q
dte−1
(q − 1)(t − (dte − 1)) + q
dte−1
− 1 1 < t ≤ n − 1
q
n−1
(q − 1)(t − (n − 1)) + q
n−1
− 1 t > n − 1
.
Then we have
G
t
(L
n
/K) = G
ψ(L
n
/K)(t)
(L
n
/K),
which gives the desired by the previous theorem.
So we know that
Art
−1
K
(G
t
(L
n
/K)) =
(
U
dte
K
/U
(n)
K
−1 ≤ t ≤ n
1 t ≥ n
.
Corollary. When t > −1, we have
G
t
(K
ab/K
) = Gal(K
ab
/K
ur
L
dte
),
and
Art
−1
K
(G
t
(K
ab
/K)) = U
(dte)
.
Proof.
Recall the following fact from the examples class: If
L/K
is finite un
ramified and
M/K
is finite totally ramified, then
LM/L
is totally ramified, and
Gal(LM/L)
∼
=
Gal(M/K) by restriction, and
G
t
(LM/K)
∼
=
G
t
(M/K).
via this isomorphism (for t > −1).
Now let
K
m
/K
be the unramified extension of degree
m
. By the lemma and
the previous corollary, we have
G
t
(K
m
L
n
/K)
∼
=
G
t
(L
n
/K) =
(
Gal(L
n
/L
dte
) −1 < t ≤ n
1 t ≥ n
=
(
Gal(K
m
L
n
/K
m
L
dte
) −1 < t ≤ n
1 t ≥ n
So we have
G
t
(K
ab
/K) = G
t
(K
ur
L
∞
/K)
= lim
←−
m,n
G
t
(K
m
L
n
/K)
= lim
←−
m,n
n≥dte
Gal(K
m
L
n
/K
m
L
dte
)
= Gal(K
ur
L
∞
/K
ur
L
dte
)
= Gal(K
ab
/K
ur
L
dte
),
and
Art
−1
K
(Gal(K
ab
/K
ur
L
dte
)) = Art
−1
K
lim
←−
m,n
n≥dte
Gal(K
m
L
n
/K
m
L
dte
)
= lim
←−
m,n
n≥dte
Art
−1
K
Gal(K
m
L
n
/K
m
L
dte
)
= lim
←−
m,n
n≥dte
U
(dte)
K
U
(n)
K
= U
dte
.
Corollary.
Let
M/K
be a finite abelian extension. Then we have an isomor
phism
Art
K
:
K
×
N(M/K)
∼
=
Gal(M/K).
Moreover, for t > −1, we have
G
t
(M/K) = Art
K
U
(dte)
K
N(M/K)
N(M/K)
!
.
Proof. We have
G
t
(M/K) =
G
t
(K
ab
/K)G(K
ab
/M)
G(K
ab
/M)
= Art
U
(dte)
K
N(M/K)
N(M/K)
!
.